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Force, mass, friction question

  • Thread starter mysticbms
  • Start date
  • #1
8
0

Homework Statement


There are no numbers given, just a question of relation.

A box with mass m rest on a board. One side of the board lifts up until the box begins to slide. For this to happen, the angle theta that the board makes with the horizontal floor depends on:

m
a (g)
N
mu


Homework Equations


Fnet = -muN + mg


The Attempt at a Solution


I tried solving it in terms of the x component of W=mg.

So F=mg-f (f = muN or friction)
F=Wx - f
Wx=mg(sin theta)
Wx-muN=Wx-muWx
Wx(1-mu)
0=mg(sin theta)(1-mu)

Those were my best attempts to come up with some sort of equation for this.
 

Answers and Replies

  • #2
1,198
5
When the force due to the weight of the box pointing down the slope exactly equals the force of friction that tries to prevent motion, the angle of inclination is defined. Equate the forces.

"Wx=mg(sin theta)"
You have the force down the incline due to weight. Write an expression for the friction force.
 
  • #3
326
3

Homework Statement


There are no numbers given, just a question of relation.

A box with mass m rest on a board. One side of the board lifts up until the box begins to slide. For this to happen, the angle theta that the board makes with the horizontal floor depends on:

m
a (g)
N
mu


Homework Equations


Fnet = -muN + mg


The Attempt at a Solution


I tried solving it in terms of the x component of W=mg.

So F=mg-f (f = muN or friction)
F=Wx - f
Wx=mg(sin theta)
Wx-muN=Wx-muWx
Wx(1-mu)
0=mg(sin theta)(1-mu)

Those were my best attempts to come up with some sort of equation for this.
I am assuming that you want to find µ.

It's easy if you draw the free-body diagram, showing the mass is at rest after the board is lifted up at some angle. Also, when drawing that diagram, you should indicate the following forces:

→ normal force [force opposite to the surface]
→ forces coming from the object
 
  • #4
8
0
Thanks for the very fast responses by the way.

Here's where I'm at now.

I drew a FBD, wish I could paste here.

I'm getting N = Wy = mg(cos theta)

At Fnet=0 Wx=f on an incline.

mg(sin theta)=f=muN

mg(sin theta)=mu*mg(cos theta)

How does this last equation tell me that theta depends on mu? Could it not also depend on N?

I think I can safely eliminate m and g because those cancel out on both sides.

m
a (g)
N
mu
 
  • #5
326
3
Thanks for the very fast responses by the way.

Here's where I'm at now.

I drew a FBD, wish I could paste here.

I'm getting N = Wy = mg(cos theta)

At Fnet=0 Wx=f on an incline.

mg(sin theta)=f=muN

mg(sin theta)=mu*mg(cos theta)

How does this last equation tell me that theta depends on mu? Could it not also depend on N?

I think I can safely eliminate m and g because those cancel out on both sides.

m
a (g)
N
mu
Yes, you are correct. Then, solve for mu, which gives the general results.
 
  • #6
1,198
5
"How does this last equation tell me that theta depends on mu? Could it not also depend on N?"

If you say that theta depends on N, you are saying that it depends on itself because

N = mg cos(theta)



That is true but in final form you wind up with

mu = tan(theta)

Do you see any N's or g's in the equation? Theta is solely dependent on mu, nothing else.
 

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